How can the Maclaurin series for sin^2(x) be simplified?

[John112]

since the maclaurin series for sin x is alternating in sign (EQ1) so when you square it to get sin^{2}(x) (EQ2) the (-1)^{n} should become (-1)^{2n} (EQ3) which can be simplified down to (EQ4), but when i checked that series at wolframalpha the series was still alternating like: Why is that? So when we square it do we ignore squaring the (-1)^{n} and put that after we're done squaring the series?In the above post when I say EQ#, I'm referring to these equations

 

[micromass]

You seem to think that $\left(\sum \alpha_n\right)^2 = \sum \alpha_n^2$. This is of course false. It's the same thing as saying that $(x + y)^2 = x^2 + y^2$.

 

[John112]

You seem to think that $\left(\sum \alpha_n\right)^2 = \sum \alpha_n^2$. This is of course false. It's the same thing as saying that $(x + y)^2 = x^2 + y^2$.

I can now see why sin^{2}(x) should be alternating, but how would I then simplify this algerbracially? how would I simplify $\left(\sum \alpha_n\right)^2$ ? Is using the half angle property for sin^{2}(x) my only method?

 

[Mark44]

I can now see why sin^{2}(x) should be alternating, but how would I then simplify this algerbracially? how would I simplify $\left(\sum \alpha_n\right)^2$ ? Is using the half angle property for sin^{2}(x) my only method?

It's probably the easiest way to go.

 

[micromass]

I can now see why sin^{2}(x) should be alternating, but how would I then simplify this algerbracially? how would I simplify $\left(\sum \alpha_n\right)^2$ ? Is using the half angle property for sin^{2}(x) my only method?

The half-angle property is indeed the easiest way. But there also is a explicit way to multiply two series. This is called the Cauchy product: http://en.wikipedia.org/wiki/Cauchy_product So you can solve it with this too, but this is a lot more complicated.

Another way is to explicitely find the derivatives of $\sin^2(x)$ and see if you can find a pattern. But this is also not a very simple way to go.